Mata, Arrianne Jaye N.Prof. Eric RoqueBS CHE301/ Monday (2:00-5:00 pm)Physical Chemistry 1Solutions to Physical Chemistry 1 Problem SetBoyles Law Pressure-Volume Relationship

1.

Given: P1= 380 torrV1= 200 mL

P2= 760 torrRequired:V2= ?Solution:

V2= P1V1 = (380 torr)(200 mL) = 100 mL P2760 torr

2.

Given:P1= 720.0 torrV1= 225 mLV2= 350.0 mLRequired:P2=?Solution:

P2= P1V1 = (720.0 torr)(225 mL) = 463 torr V2350.0 torr

3.

Given:P1= 1470 psiV1= 2.25 L

P2= 14.7 psiRequired:V2= ?Solution:

V2= P1V1 = (1470 psi)(2.25 L) = 225 L P214.7 psi

4.

Given:P1= 98.7 atmV1= 73.3 mL

P2= 4.02 atmRequired:V2= ?Solution:

P2= P1V1 = (98.7 atm)(73.3 mL) = 1800 mL V24.02 atm

5.

Given:P1= 988 mm HgV1= 535 mLV2= 1.05 L= 1050 mLRequired:P2= ?Solution:

P2= P1V1 = (988 mm Hg)(535 mL) = 503 mm Hg V21050 mLCharless Law Volume-Temperature Relationship

1. Given:V1= 3.00 LT1= -20 C = 253 K T increasesT2= 27 C = 300 K V increasesRequired:V2= ?Solution:

V2= V1T2 = (3.00 L)(300 K) = 3.56 L T1253 K

2. Given:V1= 10.5 LT1= 95 C = 368 K T2= 0 C = 273 K Required:V2= ?Solution:

V2= V1T2 = (10.5 L)(273 K) = 8 L T1 368 K

3. Solution:

Given:V1= 2.10 LT1= 25 C = 298 K T2= 150 C = 423 KRequired:V2= ?Solution:V2= V1T2 = (2.10 L)(423 K) = 2.98 L T1 298 K

4. Given:V1= 692 LT1= 602 C = 875 K T2= 23 C = 296 KRequired:V2= ?Solution:

V2= V1T2 = (692 L)(296 K) = 234 L T1 875 K

5. Given:V1= 2.10 LT1= 25 C = 298 K V2= 0.750 LRequired:T2= ?Solution:

T2= V2T1 = (0.750 L)(298 K) = 106 K = -167 C V1 2.10 L

Ideal Gas LawDetermine (a) the pressure exerted by 0.0330 mol of oxygen in an 18.0-L container at 313 K, and (b) the volume occupied by 0.200 mol of nitrogen gas at 298 K and 0.980 atm.

1. Given:n = 1.00 mol NT = 244 KP = 1.00 atmn = 1.00 atm OT = 303 KV = 15 LRequired: V and PSolution:

a. We start by solving the ideal gas equation for V.

V = nRT PV = 1.00 mol x 0.0821 Latm x 244K = 20.0 L 1.00 atm molK

b. Here we solve the ideal gas equation for P.

P = nRT VP = 0.500 mol x 0.0821 Latm x 303K = 0.829 atm 15.0 L molK

2. Given:n = 0.0330 mol OT = 313 KV = 18 LN = 0.200 mol NT = 298 KP = 0.980 atmRequired: P and VSolution:

a. P = nRT V P = 0.0330 mol x 0.0821 Latm x 313K = 0.0471 atm 18.0 L molKb. V = nRT P V = 0.200 mol x 0.0821 Latm x 298K = 4.99 L 0.980 atm molK

3. Solution:

Given:R = 0.0821 LatmmolKn = 130 g NaN3 x 1 mol NaN3 x 3 mol N2 = 6 mol N2 65.0099 g NaN3 1 mol NaN3P = 1 atmT = 25 C = 298 KRequired:V = ?Solution:

V = nRT PV = 6mol N2 x 0.0821 Latm x 298K = 146.7948 L of N2 = 73.4 L of N 1.00 atm molK

4. Given:R = 0.0821 LatmmolKn = 0.40 g CH4 x 1 mol CH4 = 0.0249 mol CH4 16.04 g CH4 P = 1 atmT = 25 C = 298 KRequired: V = ?Solution:

V = nRT PV = 0.0249 mol CH4 x 0.0821 Latm x 298K = 0.61 L 1.00 atm molK

5. Given:R = 0.0821 Latmn = 2.64mol N2molKP = 0.640 atmT = 31 C = 304 K

Required: V = ?Solution:V = nRT PV = 2.64mol N2 x 0.0821 Latm x 304K = 102.95 L 0.640 atm molK

Daltons Law of Partial Pressures

1. Given:15 g halothane22.6 g O2Required:Phalothane and PO2Solution:

15.0 g x 1 mol halothane = 0.07600 mol halothane197.4 g

22.6 g x 1 mol O2 = 0.7062 mol O2 32.00 gXhalothane = 0.07600 mol halothane = 0.09720.7822 total molesXhalothane + XO2 = 1.000 = 0.0972 + XO2 XO2 = 1.000 0.0972 = 0.903Phalothane = 0.0972 x Ptotal = 0.0972 x (862 mm Hg) = 83.8 mm HgPO2 = 0.903 x Ptotal = 0.903 x (862 mm Hg) = 778 mm Hg

2. Given:V1= 500 LP1= 739 torr P2= 760 torr Required: V2= ?Solution:

V2= P1V1 = (500 mL)(739 torr) = 486 mL dry O2 P2 760 torr

3. Given: 255 torr, 228 torr, 752 torrRequired: PSolution:

255 torr + 228 torr + 752 torr = 1235 torr x 0.00131578947 atm = 1.62 atm 1 torr4. Solution:

12 g N2 x 1 mol N2 = 0.43 mol N212 g O2 x 1 mol O2 = 0.38 mol O2 28 g N2 32 g O2P = 0.43 mol x 0.0821 Latm x 298K = 4.21 atm 2.50 L molK

P = 0.38 mol x 0.0821 Latm x 298K = 3.71 atm 2.50 L molK

Ptotal = 4.21atm + 3.71 atm = 7.92 atm

5. Given: Ptotal = 720 torrP = 22 torrRequired:PSolution:

PTot= POxygen+ PWaterPOxygen= 720 torr - 22 torr = 698 torr

Combined Gas Law

1. Given:P 1 = 754 mm HgT1= 22.3 C = 295.6 K V1 = 1.00 x 106 LP2 = 76 mm HgT2= 240 KRequired:V2 = ?Solution:

V2 = P1V1T2 = (754 mm Hg)(1.00 x 106 L)(240 K) = 8.05 x 106 LP2T1 (76 mm Hg)(295.6 K)

2. Given:P1 = 710 mm HgT1= 22.5 C = 295.5 K V1 = 21 mL = 0.021 LP2 = 740 mm HgT2= 26.5 C = 299.5 KRequired:V2 = ?Solution:

V2 = P1V1T2 = (710 mm Hg)(0.021 mL)(299.5 K) = 0.02 LP2T1 (740 mm Hg)(295.5 K)

3. Given:P1 = 300 mm HgT1= -23 C = 250 K V1 = 800 mL = 0.8 LP2 = 600 mm HgT2= 227C = 500 KRequired:V2 = ?Solution:

V2 = P1V1T2 = (300 mm Hg)(0.8 L)(500 K) = 0.8 LP2T1 (600 mm Hg)(250 K)

4.Given:P 1 = 760 mm HgT1= 22 C = 295K V1 = 400 mL = 0.4 LP2 = 360 mm HgT2= 30 C = 303 KRequired:V2 = ?Solution:

V2 = P1V1T2 = (760 mm Hg)(0.4 L)(303K) = 0.87 LP2T1 (360 mm Hg)(295 K)

5.Given:P 1 = 2 atmT1= 200 KV1 = 300 LP2 = 25 atmT2= 400 KRequired:V2 = ?Solution:

V2 = P1V1T2 = (2 atm)(300 L)(400 K) = 48 LP2T1 (25 atm)(200 K)

Grahams Law

1. Given:mass of Br2 = 160 uMass of Ar = 40 uRequired:rate of Br2 and ArSolution:

RateAr = Br2 = = 2 The Ar will diffuse twice as fast as Br2RateBr2 Ar2 1

2. Given:h = 12 hrsRequired:how long the balloon will be flatSolution:

It will take the N2balloon 2.65 times longer to deflate. 12 hrs x 2.65 = 32 hrs

3. Given:MW of O2 = 32MW of H2 = 2Required: rate of O2

4. Given:H=2.016 g/mol UF=352.02 g/molRequired:ratio of effusionSolution:

Rate of effusion of H2 = 6 = = 13.2Rate of effusion of UF6 2

5. Given:rx = 0.355 x rO2Required:identity of unknown gasSolution:

Because we are told that the unknown gas is composed of homonuclear diatomic molecules, this molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.

Gay- Lussacs Law - Pressure-Temperature Relationship

1. Given:P1= 2 atmT1= 25 C = 298 K T2= 0 C = 273 K Required:P2= ?Solution:

P2= P1T2 = (2 atm)(273 K) = 1.83 atm T1 298 K

2. Given:P1= 450 mm HgT1= 22 C = 295 K P2= 760 mm HgRequired:T2= ?Solution:T2= P2T1 = (760 mm Hg)(295 K) = 498.2 K = 225.2 C P1 450 mm Hg

3. Given:P1= 725 torrT1= 20 C = 293 K T2= 40 C = 313 KRequired:P2= ?Solution:

P2= P1T2 = (725 torr)(313 K) = 498.2 K = 774 torr T1 293 K4. Given:P1= 1.2 atmT1= 25 C = 298 K T2= 100 C = 373 KRequired:P2= ?Solution:

P2= P1T2 = (1.20 atm)(373 K) = 1.50 atm T1 298 K

5. Solution:

Kelvin unit

Amagats Law of Partial Volumes

1.Given: V1 = 24 LV2 = 15 LV3 = 68 LVtot = 220 LRequired:V of NitrogenSolution:

24 L + 15 L + 68 L = 107 L220 L 107 L = 113 L

2.Given:N = 55 L, O = 36 L, H = 75 LRequired:VtotalSolution:

Vtotal = 55 L + 36 L + 75 L = 166 L

3.Given:Vtotal = 80 L VO = 20 mL = 0.02 LRequired: VHSolution:VH = 80 L 0.02 L = 79.98 L4.Given:N = 35 L, O = 27 L and H = 17 LRequired: total volumeSolution:Vtotal = 35 L + 27 L + 17 L = 79 L5. Given:n = 10 molP = 3 atmT = 450 KV1 = 4 LV2 = 20 LRequired:partial volume of the third gasSolution:V = nRT PVtotal = 10 mol x 0.0821 Latm x 450K = 123.15 L 3 atm molK123.15 L = 4 L + 20 L + V3 = 0V3 = 99.15 L